Analytic proof

Weyl's original proof (for complex semisimple Lie algebras) was analytic in nature: it famously used the unitarian trick. Specifically, one can show that every complex semisimple Lie algebra ${\displaystyle {\mathfrak {g}}}$ is the complexification of the Lie algebra of a simply connected compact Lie group ${\displaystyle K}$.^[4] (If, for example, ${\displaystyle {\mathfrak {g}}=\mathrm {sl} (n;\mathbb {C} )}$, then ${\displaystyle K=\mathrm {SU} (n)}$.) Given a representation ${\displaystyle \pi }$ of ${\displaystyle {\mathfrak {g}}}$ on a vector space ${\displaystyle V,}$ one can first restrict ${\displaystyle \pi }$ to the Lie algebra ${\displaystyle {\mathfrak {k}}}$ of ${\displaystyle K}$. Then, since ${\displaystyle K}$ is simply connected,^[5] there is an associated representation ${\displaystyle \Pi }$ of ${\displaystyle K}$. Integration over ${\displaystyle K}$ produces an inner product on ${\displaystyle V}$ for which ${\displaystyle \Pi }$ is unitary.^[6] Complete reducibility of ${\displaystyle \Pi }$ is then immediate and elementary arguments show that the original representation ${\displaystyle \pi }$ of ${\displaystyle {\mathfrak {g}}}$ is also completely reducible.

Algebraic proof 1

Let ${\displaystyle (\pi ,V)}$ be a finite-dimensional representation of a Lie algebra ${\displaystyle {\mathfrak {g}}}$ over a field of characteristic zero. The theorem is an easy consequence of Whitehead's lemma, which says ${\displaystyle V\to \operatorname {Der} ({\mathfrak {g}},V),v\mapsto \cdot v}$ is surjective, where a linear map ${\displaystyle f:{\mathfrak {g}}\to V}$ is a derivation if ${\displaystyle f([x,y])=x\cdot f(y)-y\cdot f(x)}$. The proof is essentially due to Whitehead.^[7]

Let ${\displaystyle W\subset V}$ be a subrepresentation. Consider the vector subspace ${\displaystyle L_{W}\subset \operatorname {End} (V)}$ that consists of all linear maps ${\displaystyle t:V\to V}$ such that ${\displaystyle t(V)\subset W}$ and ${\displaystyle t(W)=0}$. It has a structure of a ${\displaystyle {\mathfrak {g}}}$-module given by: for ${\displaystyle x\in {\mathfrak {g}},t\in L_{W}}$,

${\displaystyle x\cdot t=[\pi (x),t]}$.

Now, pick some projection ${\displaystyle p:V\to V}$ onto W and consider ${\displaystyle f:{\mathfrak {g}}\to L_{W}}$ given by ${\displaystyle f(x)=[p,\pi (x)]}$. Since ${\displaystyle f}$ is a derivation, by Whitehead's lemma, we can write ${\displaystyle f(x)=x\cdot t}$ for some ${\displaystyle t\in L_{W}}$. We then have ${\displaystyle [\pi (x),p+t]=0,x\in {\mathfrak {g}}}$; that is to say ${\displaystyle p+t}$ is ${\displaystyle {\mathfrak {g}}}$-linear. Also, as t kills ${\displaystyle W}$, ${\displaystyle p+t}$ is an idempotent such that ${\displaystyle (p+t)(V)=W}$. The kernel of ${\displaystyle p+t}$ is then a complementary representation to ${\displaystyle W}$. ${\displaystyle \square }$

See also Weibel's homological algebra book.

Algebraic proof 2

Whitehead's lemma is typically proved by means of the quadratic Casimir element of the universal enveloping algebra,^[8] and there is also a proof of the theorem that uses the Casimir element directly instead of Whitehead's lemma.

Since the quadratic Casimir element ${\displaystyle C}$ is in the center of the universal enveloping algebra, Schur's lemma tells us that ${\displaystyle C}$ acts as multiple ${\displaystyle c_{\lambda }}$ of the identity in the irreducible representation of ${\displaystyle {\mathfrak {g}}}$ with highest weight ${\displaystyle \lambda }$. A key point is to establish that ${\displaystyle c_{\lambda }}$ is nonzero whenever the representation is nontrivial. This can be done by a general argument ^[9] or by the explicit formula for ${\displaystyle c_{\lambda }}$.

Consider a very special case of the theorem on complete reducibility: the case where a representation ${\displaystyle V}$ contains a nontrivial, irreducible, invariant subspace ${\displaystyle W}$ of codimension one. Let ${\displaystyle C_{V}}$ denote the action of ${\displaystyle C}$ on ${\displaystyle V}$. Since ${\displaystyle V}$ is not irreducible, ${\displaystyle C_{V}}$ is not necessarily a multiple of the identity, but it is a self-intertwining operator for ${\displaystyle V}$. Then the restriction of ${\displaystyle C_{V}}$ to ${\displaystyle W}$ is a nonzero multiple of the identity. But since the quotient ${\displaystyle V/W}$ is a one dimensional—and therefore trivial—representation of ${\displaystyle {\mathfrak {g}}}$, the action of ${\displaystyle C}$ on the quotient is trivial. It then easily follows that ${\displaystyle C_{V}}$ must have a nonzero kernel—and the kernel is an invariant subspace, since ${\displaystyle C_{V}}$ is a self-intertwiner. The kernel is then a one-dimensional invariant subspace, whose intersection with ${\displaystyle W}$ is zero. Thus, ${\displaystyle \mathrm {ker} (V_{C})}$ is an invariant complement to ${\displaystyle W}$, so that ${\displaystyle V}$ decomposes as a direct sum of irreducible subspaces:

${\displaystyle V=W\oplus \mathrm {ker} (C_{V})}$.

Although this establishes only a very special case of the desired result, this step is actually the critical one in the general argument.

Algebraic proof 3

The theorem can be deduced from the theory of Verma modules, which characterizes a simple module as a quotient of a Verma module by a maximal submodule.^[10] This approach has an advantage that it can be used to weaken the finite-dimensionality assumptions (on algebra and representation).

Let ${\displaystyle V}$ be a finite-dimensional representation of a finite-dimensional semisimple Lie algebra ${\displaystyle {\mathfrak {g}}}$ over an algebraically closed field of characteristic zero. Let ${\displaystyle {\mathfrak {b}}={\mathfrak {h}}\oplus {\mathfrak {n}}_{+}\subset {\mathfrak {g}}}$ be the Borel subalgebra determined by a choice of a Cartan subalgebra and positive roots. Let ${\displaystyle V^{0}=\{v\in V|{\mathfrak {n}}_{+}(v)=0\}}$. Then ${\displaystyle V^{0}}$ is an ${\displaystyle {\mathfrak {h}}}$-module and thus has the ${\displaystyle {\mathfrak {h}}}$-weight space decomposition:

${\displaystyle V^{0}=\bigoplus _{\lambda \in L}V_{\lambda }^{0}}$

where ${\displaystyle L\subset {\mathfrak {h}}^{*}}$. For each ${\displaystyle \lambda \in L}$, pick ${\displaystyle 0\neq v_{\lambda }\in V_{\lambda }}$ and ${\displaystyle V^{\lambda }\subset V}$ the ${\displaystyle {\mathfrak {g}}}$-submodule generated by ${\displaystyle v_{\lambda }}$ and ${\displaystyle V'\subset V}$ the ${\displaystyle {\mathfrak {g}}}$-submodule generated by ${\displaystyle V^{0}}$. We claim: ${\displaystyle V=V'}$. Suppose ${\displaystyle V\neq V'}$. By Lie's theorem, there exists a ${\displaystyle {\mathfrak {b}}}$-weight vector in ${\displaystyle V/V'}$; thus, we can find an ${\displaystyle {\mathfrak {h}}}$-weight vector ${\displaystyle v}$ such that ${\displaystyle 0\neq e_{i}(v)\in V'}$ for some ${\displaystyle e_{i}}$ among the Chevalley generators. Now, ${\displaystyle e_{i}(v)}$ has weight ${\displaystyle \mu +\alpha _{i}}$. Since ${\displaystyle L}$ is partially ordered, there is a ${\displaystyle \lambda \in L}$ such that ${\displaystyle \lambda \geq \mu +\alpha _{i}}$; i.e., ${\displaystyle \lambda >\mu }$. But this is a contradiction since ${\displaystyle \lambda ,\mu }$ are both primitive weights (it is known that the primitive weights are incomparable.). Similarly, each ${\displaystyle V^{\lambda }}$ is simple as a ${\displaystyle {\mathfrak {g}}}$-module. Indeed, if it is not simple, then, for some ${\displaystyle \mu <\lambda }$, ${\displaystyle V_{\mu }^{0}}$ contains some nonzero vector that is not a highest-weight vector; again a contradiction. ${\displaystyle \square }$